# Limit of function $f(x) = \sqrt{(xa + d)^2 + x^2 b^2} - \sqrt{(xa - d)^2 + x^2 b^2}$

I tried to calculate limit when $$x$$ goes to infinity for the following function $$f(x) = \sqrt{(xa + d)^2 + x^2 b^2} - \sqrt{(xa - d)^2 + x^2 b^2}$$ where $$a$$, $$b$$, $$d$$ are some positive constants.

It's easy to see that terms before and after minus sign goes to infinity so that gives me indeterminate symbol. Is there some way to solve this problem?

• What methods are you allowed to use? – Yuriy S Sep 6 '19 at 13:54
• You could multiply by $\sqrt{(xa + d)^2 + x^2 b^2} + \sqrt{(xa - d)^2 + x^2 b^2} / \sqrt{(xa + d)^2 + x^2 b^2} + \sqrt{(xa - d)^2 + x^2 b^2}$ – Milan Sep 6 '19 at 13:55
• @YuriyS Any methoh, this is my private investigation out of curiosity -- not a homework. – Trismegistos Sep 6 '19 at 13:59
• @Milan What would that give me? – Trismegistos Sep 6 '19 at 14:00
• @Trismegistos You could try it and find out for yourself what it gives you... – David C. Ullrich Sep 6 '19 at 14:51

Multiplying by $$\sqrt{\cdot} + \sqrt{\cdot}$$ at numerator and denomitor your get

\begin{aligned}f(x)&= \frac{4adx}{{\sqrt{(xa + d)^2 + x^2 b^2} + \sqrt{(xa - d)^2 + x^2 b^2}}}\\ \end{aligned}

And therefore $$\lim\limits_{x \to \infty} f(x) = \frac{2ad}{\sqrt{a^2+b^2}}$$ by pulling $$x$$ at the denominator as

\begin{aligned}\sqrt{(xa + d)^2 + x^2 b^2} + \sqrt{(xa - d)^2 + x^2 b^2}&=x\left(\sqrt{(a+ d/x)^2 + b^2} + \sqrt{(a - d/x)^2 + b^2}\right)\\ \end{aligned} for $$x>0$$ and $$\lim\limits_{x \to \infty} d/x =0$$.

Easy then to get $$\lim\limits_{x \to -\infty} f(x)$$ as $$f$$ is odd.

• I don't understand how you get such simple form in denominator I tried to work it out but I got $\sqrt{x^2 a^2 + 2xad + d^2 + x^2 b^2} - \sqrt{x^2 a^2 - 2xad + d^2 + x^2 b^2}$ I can pull $x^{-2}$ into square root but because of 2xad term I'm still left with $x^{-1}$ under square root. – Trismegistos Sep 7 '19 at 8:08
• I added some intermediate steps. – mathcounterexamples.net Sep 7 '19 at 8:19
• Thank you. Multiplying by conjugate was good trick. I wonder how did you come up with such trick. – Trismegistos Sep 7 '19 at 8:25
• Classical when you have a sum or difference of square roots. – mathcounterexamples.net Sep 7 '19 at 8:32

The simplest solution is probably the one in mathcounterexamples.net's answer.

But that's just an algebraic trick. You can also give a solution using calculus:

If $$0 there exists $$\xi\in(s,t)$$ such that $$\sqrt t-\sqrt s=\frac{t-s}{2\sqrt\xi}$$.

Proof: Apply the Mean Value Theorem to the function $$f(x)=\sqrt x$$.

Now let $$t = (xa+d)^2+x^2b^2$$ and $$s=(xa-d)^2+x^2b^2$$ and see what happens...

(One reason to do this is that this method can be applied in other situations where the algebraic trick is not available. MVT is the source of all wisdom.)

Details added in response to a comment: Let's say $$a$$, $$x$$ annd $$d$$ are positive, so if we define $$s$$ and $$t$$ as above we have $$s. Now $$s<\xi leads to $$\frac{t-s}{2\sqrt t}<\frac{t-s}{2\sqrt \xi}<\frac{t-s}{2\sqrt s},$$or in other words $$\frac{2xad}{\sqrt{(xa+d)^2+x^2b^2}}<\frac{t-s}{2\sqrt\xi}<\frac{2xad}{\sqrt{(xa-d)^2+x^2b^2}},$$or$$\frac{2ad}{\sqrt{(a+d/x)^2+b^2}}<\frac{t-s}{2\sqrt\xi}<\frac{2ad}{\sqrt{(a-d/x)^2+b^2}}.$$Now let $$x\to\infty$$ and you're done.

(Hehe: When I posted this answer I hadn't worked out those details; it just seemed clear that MVT "must" tell us what we need to know about $$f(t)-f(s)$$. Sure enough, MVT is the key to understanding the universe.)

## The Moral

Any time you need to figure out something about a quantity of the form $$f(t)-f(s)$$ (where $$f$$ is differentiable) you should always consider applying MVT. Because MVT always tells you something about $$f(t)-f(s)$$, and if $$t-s$$ is not too big that "something" will often include everything there is to know about $$f(t)-f(s)$$. (Not going to give a precise definition of "not too big" here - work out what MVT tells you, and if that's what you needed to know then $$t-s$$ was not too big...)

• +1, because now I know about the source of all wisdom. – Michael Hoppe Sep 6 '19 at 16:58
• It's really nice, I have never seen MVT in this context :) – Botond Sep 6 '19 at 17:26
• Using this method result is $2xad / \sqrt{\xi}$ which makes limit goes to infinity but my experiments with wolfram alpha shows that limit is finite. So I'm not sure about result... – Trismegistos Sep 7 '19 at 7:42
• @Trismegistos I don't know why you think $2xad/\sqrt\xi$ tends to infinity, but it's not so. See revised version above... – David C. Ullrich Sep 7 '19 at 13:02
• @MichaelHoppe It's true. Yesterday when I posted this answer I didn't know for a fact that the method would work, I simply believed it "must" work because MVT so often tells us everything there is to know. Worked it out when the OP expressed skepticism, and sure enough. (See The Moral above...) – David C. Ullrich Sep 7 '19 at 13:05

For $$x\not=0$$, \begin{align}\sqrt{(xa \pm d)^2 + x^2 b^2} &=|x|\sqrt{\left(a \pm \frac{d}{x}\right)^2+ b^2}\\ &=|x|\sqrt{a^2+b^2}\left(1\pm\frac{2ad}{(a^2+b^2)x}+o(1/x)\right)^{1/2}\\ &=|x|\sqrt{a^2+b^2}\left(1\pm\frac{ad}{(a^2+b^2)x}+o(1/x)\right)\\ &=|x|\sqrt{a^2+b^2}\pm\frac{ad|x|/x}{\sqrt{a^2+b^2}}+o(1) \end{align} where we used the expansion at $$t=0$$, $$(1+t)^{1/2}=1+t/2+o(t)$$.

Now it should be easy to verify that $$\lim\limits_{x \to +\infty} f(x) = \frac{2ad}{\sqrt{a^2+b^2}}\quad\text{and}\quad \lim\limits_{x \to -\infty} f(x) = -\frac{2ad}{\sqrt{a^2+b^2}}.$$

Let $$t=1/(2 d x)$$. Then the limit of interest can be written as

$$\lim_{t\to 0^{\pm}}\left(\sqrt{(2a d/t + d)^2 + b^2 d^2/t^2} - \sqrt{(2a d/t - d)^2 + b^2 d^2/t^2}\right)\\ \hspace{4cm}=2d\,\text{sgn}(t)\times \lim_{t\to 0^{\pm}}\frac{1}{t}\left[\sqrt{(a+t/2)^2+b^2}-\sqrt{(a-t/2)^2+b^2}\right].$$ The definition of the derivative therefore yields the limits for $$t\to 0^{\pm}$$ as

$$\pm 2d \frac{d}{dt}\left( \sqrt{(a+t)^2+b^2}\right)_{t=0}=\pm \frac{2ad}{\sqrt{a^2+b^2}}.$$